The derivative of the sine of the angle is equal to the cosine of the same angle

Given the simplest trigonometry function y = Sin (x), it is differentiable at each of its points from the entire domain of definition. It is necessary to prove that the derivative of the sine of any argument is equal to the cosine of the same angle, that is, y '= Cos (x).

The proof is based on the definition of the derivative of the function

We define x (arbitrary) in some small neighborhood Δx of a particular point x0. Let us show the value of the function in it and at the point x to find the increment of a given function. If Δx is the increment of the argument, then the new argument is x ₀ + Δx = x, the value of this function for a given value of the argument y (x) is Sin (x ₀ + Δx), the value of the function at a particular point y (x ₀ ) .

Now we have Δy = Sin (x ₀ + Δx) -Sin (x ₀ ) is the increment of the function obtained.

By the sine formula of the sum of two unequal angles, we will transform the difference Δy.

(Cos) = cos (Δx) + Cos (x ₀ ) · Sin (Δx) minus Sin (x ₀ ) = (Cos (Δx) -1) · Sin (x ₀ ) + Cos (x ₀ ) · Sin (Δx).

Performed a permutation of the terms, grouped the first with the third Sin (x ₀ ), carried a common multiplier - sine - for the brackets. We obtained in the expression the difference Cos (Δx) -1. It remains to change the sign in front of the bracket and in parentheses. Knowing which is 1-Cos (Δx), we make a substitution and obtain a simplified expression Δy, which we then divide by Δx.
Δy / Δx will have the form: Cos (x ₀ ) · Sin (Δx) / Δx-2 · Sin ² (0.5 · Δx) · Sin (x ₀ ) / Δx. This is the ratio of the increment of the function to the allowed increment of the argument.

It remains to find the limit of the ratio lim obtained for Δx tending to zero.

It is known that the limit Sin (Δx) / Δx is equal to 1, under this condition. The expression 2 · Sin ² (0,5 · Δx) / Δx in the resulting quotient is reduced to the product containing the first remarkable limit as a multiplier: divide the numerator and denominator of the fraction by 2, replace the square of the sine by the product. Like this:
(Sin (0.5 · Δx) / (0.5 · Δx)) · Sin (Δx / 2).
The limit of this expression for Δx tending to zero is equal to zero (1 multiplied by 0). It turns out that the limit of the ratio Δy / Δx is Cos (x ₀ ) · 1-0, this is Cos (x ₀ ), an expression that does not depend on Δx tending to 0. This leads to the conclusion that the sine derivative of any angle x is Cosine x, we write as y '= Cos (x).

The resulting formula is entered in the known table of derivatives, where all the elementary functions

When solving problems where the derivative of a sinus occurs, one can use the rules of differentiation and ready-made formulas from the table. For example: find the derivative of the simplest function y = 3 · Sin (x) -15. We use the elementary rules of differentiation, the removal of the numerical factor behind the sign of the derivative, and the calculation of the derivative of a constant number (it is zero). We apply the tabulated value of the derivative of the sine of the angle x, equal to Cos (x). We get the answer: y '= 3 · Cos (x) -O. This derivative, in turn, is also an elementary function y = 3 · Cos (x).

The derivative of the sine is squared from any argument

When calculating this expression (Sin ² (x)) ', it is necessary to remember how the complex function is differentiated. So, y = sin ² (x) - is a power function, since the sine is squared. Its argument is also a trigonometric function, Complex argument. The result in this case is equal to the product whose first factor is the derivative of the square of the given complex argument, and the second is the derivative of the sine. This is how the rule for differentiating a function of a function looks like: (u (v (x))) 'is equal to (u (v (x)))' (v (x)) '. The expression v (x) is a complex argument (internal function). If the function "igrok is equal to the sine in the square x" is given, then the derivative of this complex function is y '= 2 · Sin (x) · Cos (x). In the product, the first doubled multiplier is the derivative of the known power function, and Cos (x) is the derivative of the sine, the argument of a complex quadratic function. The final result can be transformed by using the trigonometric sine formula of the double angle. Answer: the derivative is Sin (2 · x). This formula is easily remembered, it is often used as a tabular.